[proofplan] We construct the absolutely continuous and singular parts using a supremum argument. Define $\mathcal{S} = \{ \rho : \rho \text{ is a measure on } (X, \mathcal{A}), \rho \le \nu, \rho \ll \mu \}$ and set $\nu_{\mathrm{ac}}$ to be the supremum of this family. The construction proceeds by showing that countable suprema of absolutely continuous measures remain absolutely continuous, extracting a maximizing sequence, and verifying that the remainder $\nu_s := \nu - \nu_{\mathrm{ac}}$ is singular with respect to $\mu$. Uniqueness follows from the observation that any other decomposition would produce an absolutely continuous measure exceeding the supremum. [/proofplan]

[step:Define the family $\mathcal{S}$ and extract a maximizing sequence]Let $(X_k)_{k=1}^\infty$ be an exhaustion of $X$ with $X_k \uparrow X$ and $\nu(X_k) < \infty$ (such an exhaustion exists because $\nu$ is $\sigma$-finite). Define: \begin{align*} \mathcal{S} := \{ \rho : \rho \text{ is a measure on } (X, \mathcal{A}), \; \rho(A) \le \nu(A) \text{ for all } A \in \mathcal{A}, \; \rho \ll \mu \}. \end{align*} The zero measure belongs to $\mathcal{S}$, so $\mathcal{S} \neq \varnothing$. For each $k \in \mathbb{N}$, define: \begin{align*} s_k := \sup \{ \rho(X_k) : \rho \in \mathcal{S} \}. \end{align*} Since $\rho(X_k) \le \nu(X_k) < \infty$ for every $\rho \in \mathcal{S}$, each $s_k$ is finite. For each $k$, choose a sequence $(\rho_{k,j})_{j=1}^\infty$ in $\mathcal{S}$ with $\rho_{k,j}(X_k) \to s_k$ as $j \to \infty$.[/step]

[guided]The idea is to build $\nu_{\mathrm{ac}}$ as the largest measure that is simultaneously $\le \nu$ and absolutely continuous with respect to $\mu$. The supremum must be taken carefully because we are maximizing over a family of measures, not over real numbers. The strategy is to work on the exhaustion sets $X_k$ (where $\nu$ is finite) and take suprema of real numbers $\rho(X_k)$, then combine via a diagonal argument. The $\sigma$-finiteness of $\nu$ is essential: it ensures $s_k < \infty$ so the suprema are finite and maximizing sequences exist. Without $\sigma$-finiteness, we could have $s_k = \infty$, and the construction would break down.[/guided]

[step:Close $\mathcal{S}$ under finite maxima to build a single maximizing sequence] [claim:Closure of $\mathcal{S}$ under pairwise maxima] If $\rho_1, \rho_2 \in \mathcal{S}$, then the measure $\rho_1 \vee \rho_2$ defined by $(\rho_1 \vee \rho_2)(A) := \sup \{ \rho_1(B) + \rho_2(A \setminus B) : B \in \mathcal{A}, B \subseteq A \}$ belongs to $\mathcal{S}$. [/claim] [proof] We verify countable additivity. Let $(A_m)_{m=1}^\infty$ be pairwise disjoint sets in $\mathcal{A}$ with $A = \bigcup_{m=1}^\infty A_m$. For any $B \subseteq A$ with $B \in \mathcal{A}$, set $B_m := B \cap A_m$. Then $(B_m)$ are disjoint, $B = \bigcup_m B_m$, and: \begin{align*} \rho_1(B) + \rho_2(A \setminus B) &= \sum_{m=1}^\infty \rho_1(B_m) + \sum_{m=1}^\infty \rho_2(A_m \setminus B_m) = \sum_{m=1}^\infty [\rho_1(B_m) + \rho_2(A_m \setminus B_m)] \\ &\le \sum_{m=1}^\infty (\rho_1 \vee \rho_2)(A_m). \end{align*} Taking the supremum over $B$ gives $(\rho_1 \vee \rho_2)(A) \le \sum_m (\rho_1 \vee \rho_2)(A_m)$. For the reverse inequality, for each $m$ and $\varepsilon > 0$, choose $B_m \subseteq A_m$ achieving $\rho_1(B_m) + \rho_2(A_m \setminus B_m) \ge (\rho_1 \vee \rho_2)(A_m) - \varepsilon/2^m$. Setting $B = \bigcup_m B_m \subseteq A$: \begin{align*} (\rho_1 \vee \rho_2)(A) \ge \rho_1(B) + \rho_2(A \setminus B) = \sum_{m=1}^\infty [\rho_1(B_m) + \rho_2(A_m \setminus B_m)] \ge \sum_{m=1}^\infty (\rho_1 \vee \rho_2)(A_m) - \varepsilon. \end{align*} Since $\varepsilon > 0$ is arbitrary, countable additivity holds. Next, $(\rho_1 \vee \rho_2)(A) \le \nu(A)$: for any $B \subseteq A$, $\rho_1(B) + \rho_2(A \setminus B) \le \nu(B) + \nu(A \setminus B) = \nu(A)$. Finally, $(\rho_1 \vee \rho_2) \ll \mu$: if $\mu(A) = 0$, then the only $B \subseteq A$ with $B \in \mathcal{A}$ satisfies $\mu(B) = 0$, so $\rho_1(B) = 0$ and $\rho_2(A \setminus B) = 0$ (since $\mu(A \setminus B) \le \mu(A) = 0$). Thus $(\rho_1 \vee \rho_2)(A) = 0$. [/proof] For each $k$, iteratively replace pairs in the maximizing sequence $(\rho_{k,j})_j$ by their pairwise maxima. Define $\sigma_{k,1} := \rho_{k,1}$ and $\sigma_{k,j} := \sigma_{k,j-1} \vee \rho_{k,j}$ for $j \ge 2$. Then $\sigma_{k,j} \in \mathcal{S}$, $\sigma_{k,j}(X_k) \ge \rho_{k,j}(X_k)$, and the sequence $(\sigma_{k,j}(X_k))_j$ is nondecreasing with limit $s_k$. Now define a single sequence that simultaneously maximizes on every $X_k$. By a diagonal construction, define $\tau_j := \sigma_{1,j} \vee \sigma_{2,j} \vee \cdots \vee \sigma_{j,j}$. Then $\tau_j \in \mathcal{S}$ (by iterated application of the claim), $(\tau_j(A))_j$ is nondecreasing for every $A \in \mathcal{A}$, and $\tau_j(X_k) \to s_k$ for every $k$. [/step]

[step:Define $\nu_{\mathrm{ac}}$ as the pointwise limit and verify it is a measure in $\mathcal{S}$]Define: \begin{align*} \nu_{\mathrm{ac}}(A) := \lim_{j \to \infty} \tau_j(A) = \sup_{j} \tau_j(A) \quad \text{for every } A \in \mathcal{A}. \end{align*} The limit exists (in $[0, \infty]$) because $(\tau_j(A))_j$ is nondecreasing. We verify that $\nu_{\mathrm{ac}}$ is a measure: **Empty set:** $\nu_{\mathrm{ac}}(\varnothing) = \lim_j \tau_j(\varnothing) = 0$. **Countable additivity:** Let $(A_m)_{m=1}^\infty$ be pairwise disjoint in $\mathcal{A}$. For each $j$, $\tau_j\left(\bigcup_m A_m\right) = \sum_m \tau_j(A_m)$. Taking $j \to \infty$ on both sides and applying the [Monotone Convergence Theorem](/theorems/509) (for series: interchange of $\lim_j$ and $\sum_m$ when all terms are nonneg and nondecreasing in $j$): \begin{align*} \nu_{\mathrm{ac}}\left(\bigcup_{m=1}^\infty A_m\right) = \lim_{j \to \infty} \sum_{m=1}^\infty \tau_j(A_m) = \sum_{m=1}^\infty \lim_{j \to \infty} \tau_j(A_m) = \sum_{m=1}^\infty \nu_{\mathrm{ac}}(A_m). \end{align*} **Bound:** $\nu_{\mathrm{ac}}(A) = \lim_j \tau_j(A) \le \nu(A)$ since $\tau_j(A) \le \nu(A)$ for every $j$. **Absolute continuity:** If $\mu(A) = 0$, then $\tau_j(A) = 0$ for every $j$ (since $\tau_j \ll \mu$), so $\nu_{\mathrm{ac}}(A) = 0$. Hence $\nu_{\mathrm{ac}} \ll \mu$. Thus $\nu_{\mathrm{ac}} \in \mathcal{S}$, and $\nu_{\mathrm{ac}}(X_k) = s_k$ for every $k$.[/step]

[guided]The interchange of limit and summation uses the following form of monotone convergence: if $a_{j,m} \ge 0$ and $a_{j,m}$ is nondecreasing in $j$ for each fixed $m$, then $\lim_j \sum_m a_{j,m} = \sum_m \lim_j a_{j,m}$. Here $a_{j,m} = \tau_j(A_m)$, which is nondecreasing in $j$ (since $\tau_{j+1} \ge \tau_j$ pointwise on $\mathcal{A}$) and nonneg. This is the [Monotone Convergence Theorem](/theorems/509) applied to counting measure on $\mathbb{N}$. The fact that $\nu_{\mathrm{ac}}(X_k) = s_k$ follows from $\tau_j(X_k) \to s_k$ (by construction of the diagonal sequence).[/guided]

[step:Define $\nu_s := \nu - \nu_{\mathrm{ac}}$ and verify it is singular with respect to $\mu$]Since $\nu_{\mathrm{ac}}(A) \le \nu(A)$ for every $A \in \mathcal{A}$, the set function $\nu_s := \nu - \nu_{\mathrm{ac}}$ is a well-defined nonneg measure on $(X, \mathcal{A})$ (countable additivity follows from that of $\nu$ and $\nu_{\mathrm{ac}}$). We show $\nu_s \perp \mu$. Let $(Y_k)_{k=1}^\infty$ be an exhaustion of $X$ with $Y_k \uparrow X$ and $\mu(Y_k) < \infty$ (using the $\sigma$-finiteness of $\mu$). We may assume $Y_k = X_k$ by replacing the exhaustion with $X_k \cap Y_k$ if necessary, and relabelling. Suppose for contradiction that $\nu_s \not\perp \mu$. Then there exists $k_0$ such that for every $A \in \mathcal{A}$ with $\mu(A) = 0$, we have $\nu_s(X_{k_0} \setminus A) > 0$ — in other words, $\nu_s|_{X_{k_0}}$ is not carried by a $\mu$-null set. This means $\nu_s|_{X_{k_0}}$ is not singular with respect to $\mu|_{X_{k_0}}$. Since $\nu_s|_{X_{k_0}}$ and $\mu|_{X_{k_0}}$ are finite measures that are not mutually singular, there exists $\varepsilon > 0$ and a set $E \in \mathcal{A}$ with $E \subseteq X_{k_0}$ such that $\nu_s(A) \ge \varepsilon \mu(A)$ for every $A \in \mathcal{A}$ with $A \subseteq E$, and $\mu(E) > 0$. To find such $E$ and $\varepsilon$: consider the signed measure $\nu_s|_{X_{k_0}} - \frac{1}{m} \mu|_{X_{k_0}}$ for $m \in \mathbb{N}$. If $\nu_s|_{X_{k_0}} \perp \mu|_{X_{k_0}}$, there is nothing to prove. Otherwise, for some $m$, the signed measure $\nu_s - \frac{1}{m}\mu$ restricted to $X_{k_0}$ has a positive set $E$ with $\mu(E) > 0$. (If every positive set had $\mu$-measure zero for every $m$, then for each $m$, $\nu_s(A) \le \frac{1}{m}\mu(A)$ for all $A \subseteq X_{k_0}$ with $A$ lying in the negative set. Taking $m \to \infty$ would give $\nu_s = 0$ on $X_{k_0}$ modulo a $\mu$-null set, contradicting our assumption.) Set $\varepsilon := 1/m$. Now define $\rho: \mathcal{A} \to [0, \infty)$ by $\rho(A) := \varepsilon \mu(A \cap E)$. Then $\rho \ll \mu$ and $\rho(A) = \varepsilon \mu(A \cap E) \le \nu_s(A \cap E) \le \nu_s(A)$ for every $A \in \mathcal{A}$. Consider $\nu_{\mathrm{ac}} + \rho$. For every $A \in \mathcal{A}$: \begin{align*} (\nu_{\mathrm{ac}} + \rho)(A) = \nu_{\mathrm{ac}}(A) + \rho(A) \le \nu_{\mathrm{ac}}(A) + \nu_s(A) = \nu(A). \end{align*} Also $\nu_{\mathrm{ac}} + \rho \ll \mu$ since both $\nu_{\mathrm{ac}} \ll \mu$ and $\rho \ll \mu$. Hence $\nu_{\mathrm{ac}} + \rho \in \mathcal{S}$. But: \begin{align*} (\nu_{\mathrm{ac}} + \rho)(X_{k_0}) = \nu_{\mathrm{ac}}(X_{k_0}) + \varepsilon \mu(E) = s_{k_0} + \varepsilon \mu(E) > s_{k_0}, \end{align*} contradicting the definition of $s_{k_0}$ as the supremum of $\rho(X_{k_0})$ over $\rho \in \mathcal{S}$. Therefore $\nu_s \perp \mu$.[/step]

[guided]The key step is showing that the remainder $\nu_s$ must be singular. The argument proceeds by contradiction: if $\nu_s$ is not singular with respect to $\mu$, then $\nu_s$ has a nontrivial absolutely continuous component, which we can extract and add to $\nu_{\mathrm{ac}}$ — violating the maximality of $\nu_{\mathrm{ac}}$. The extraction uses a Hahn-decomposition-style argument for the signed measure $\nu_s - \varepsilon \mu$ on a finite-measure set. If $\nu_s$ is not singular with respect to $\mu$ on $X_{k_0}$, then for some $m \in \mathbb{N}$, the signed measure $\nu_s - \frac{1}{m}\mu$ restricted to $X_{k_0}$ has a positive set $E$ with $\mu(E) > 0$. The measure $\rho(A) = \frac{1}{m}\mu(A \cap E)$ is then absolutely continuous and dominated by $\nu_s$, so $\nu_{\mathrm{ac}} + \rho \in \mathcal{S}$. This contradicts the maximality of $s_{k_0}$ because $(\nu_{\mathrm{ac}} + \rho)(X_{k_0}) = s_{k_0} + \frac{1}{m}\mu(E) > s_{k_0}$.[/guided]

[step:Prove uniqueness of the decomposition]Suppose $\nu = \nu_{\mathrm{ac}}' + \nu_s'$ is another decomposition with $\nu_{\mathrm{ac}}' \ll \mu$ and $\nu_s' \perp \mu$. Then $\nu_{\mathrm{ac}}' \le \nu$ and $\nu_{\mathrm{ac}}' \ll \mu$, so $\nu_{\mathrm{ac}}' \in \mathcal{S}$, which gives $\nu_{\mathrm{ac}}'(X_k) \le s_k = \nu_{\mathrm{ac}}(X_k)$ for every $k$. Now $\nu_{\mathrm{ac}} - \nu_{\mathrm{ac}}' = \nu_s' - \nu_s$. The left side is absolutely continuous with respect to $\mu$ (as the difference of two absolutely continuous measures). The right side: let $X = S_1 \cup S_1^c$ witness $\nu_s \perp \mu$ (with $\nu_s(S_1^c) = 0$ and $\mu(S_1) = 0$) and $X = S_2 \cup S_2^c$ witness $\nu_s' \perp \mu$ (with $\nu_s'(S_2^c) = 0$ and $\mu(S_2) = 0$). Then $\nu_s' - \nu_s$ is carried on $S_1 \cup S_2$ (since both $\nu_s$ and $\nu_s'$ vanish outside $S_1 \cup S_2$), and $\mu(S_1 \cup S_2) \le \mu(S_1) + \mu(S_2) = 0$. So $\nu_s' - \nu_s$ is singular with respect to $\mu$. A signed measure that is both absolutely continuous and singular with respect to $\mu$ must be zero: if $\sigma \ll \mu$ and $\sigma \perp \mu$, let $X = T \cup T^c$ witness $\sigma \perp \mu$ with $\mu(T) = 0$. Then for any $A \in \mathcal{A}$, $\sigma(A) = \sigma(A \cap T) + \sigma(A \cap T^c) = \sigma(A \cap T)$. But $\mu(A \cap T) \le \mu(T) = 0$, so $\sigma(A \cap T) = 0$ by absolute continuity. Hence $\sigma \equiv 0$. Therefore $\nu_{\mathrm{ac}} - \nu_{\mathrm{ac}}' = 0$, i.e., $\nu_{\mathrm{ac}} = \nu_{\mathrm{ac}}'$, and consequently $\nu_s = \nu_s'$.[/step]

[guided]The uniqueness argument has a clean structure. Any alternative absolutely continuous part $\nu_{\mathrm{ac}}'$ belongs to $\mathcal{S}$, so it cannot exceed $\nu_{\mathrm{ac}}$ on the exhaustion sets $X_k$. The difference $\nu_{\mathrm{ac}} - \nu_{\mathrm{ac}}' = \nu_s' - \nu_s$ is simultaneously absolutely continuous (from the left side) and singular (from the right side) with respect to $\mu$. The general principle that a measure which is both absolutely continuous and singular must vanish is a fundamental dichotomy in measure theory. The proof is direct: if $\sigma \ll \mu$ and $\sigma$ is carried on a $\mu$-null set $T$, then $\sigma(A) = \sigma(A \cap T)$, and $\mu(A \cap T) = 0$ forces $\sigma(A \cap T) = 0$ by absolute continuity.[/guided]